Question

请问 (x^2 - 1)开根号 的倒数 dx 怎么求?

Answer 1

设1/(x^2-1)=1/(x+1)(x-1)

=a/(x+1)+b/(x-1)

=[(a+b)x+(b-a)]/(x+1)(x-1)

所以a+b=0,b-a=1

解得a=-1/2,b=1/2

所以∫1/√(x^2 - 1) dx=-1/2∫1/(x+1)dx+1/2∫1/(x-1)dx

=-1/2*ln|x+1|+1/2*ln|x-1|+C

=1/2*ln|(x-1)/(x+1)|+C